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2009/10 AHSMC Part 1 Problem 16

Problem 16 of 16

Suppose that 2−992 - \sqrt{99} is a root of x2+ax+bx^2 + ax + b where bb is a negative real number and aa is an integer. The largest possible value of aa is

Answer choices

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Solution

Let the other root be tt. Then −a=t+2−99-a = t + 2 - \sqrt{99}. Now b=t(2−99)<0b = t(2 - \sqrt{99}) < 0, so that t=99−2−a>0t = \sqrt{99} - 2 - a > 0. Since aa is an integer, a≤9−2=7a \le 9 - 2 = 7. The answer is (C).

University of Alberta